Thank you very much, chusub1. I don't know if I will have a post for Georgia 4/14 Evening, yet. I am on vacation and my wife and I have plans for today. We shall see if I'm home when I need to compute it. Each one of my posts are based on the last draw, so I can't do it until the Georgia Midday has happened.

I am new to the group and need to learn. I also live in Georgia, as the other guy that posted in this link. I would like to learn how you create the list of numbers and if you use a program or certain formula, I would appreciate knowing what or how you do it. Also, of the 40 combinations, do you play all of them? Sorry, to seem dumb, but I need to learn.

Thanks in advance for your help!!! I hope you and your wife are having a great time on your vacation!

No problem, TeddyBear. I start by looking at the past draws and write down the numbers in the sequence they were drawn. I will end up with all ten digits in a row (or 2 columns of 5). I then decide which number or number is going to be my Key # (I will explain later), and then I play 1 number from "Column A" and 2 numbers from "Column B" or vice versa.

Looking at recent draws in Georgia:

04-14 eve

05278-39164

04-14 mid

052

7

-1

78350-91264

548

2-1

04-13 eve

783

18

-3

53091-26487

092

1-2

04-13 mid

533

11

+1

09125-63487

(577)

(1-2)

04-12 eve

091

10

-2

20563-49871

270

1-2

04-12 mid

200

2

+1

50634-29871

(622)

(2-1)

04-11 eve

506

11

+4

30429-58671

628

1-2

04-10 eve

304

7

+2

03249-58671

214

3-0

04-10 mid

032

5

+3

49586-03721

679

0-3

04-09 eve

499

22

0

58960-43721

(633)

(2-1)

The numbers that are highlighted in pink are doubles. The third column is the sum of the digits and it is yellow if a neighbor (2 numbers in sequence) is in that draw. The next column is the "jump" in the "sum column" from the last draw's sum. I say sum column because I'm thinking of my chart (well, not mine!!!) with the 220 unique numbers divided into 10 equal columns with each column numbered with just the ones digit of the sum. So all numbers that total 2, 12, or 22 would be in the same column. That would be 10 columns of 22 numbers each.

The sum jump column is green if the number drawn has all even or all odd digits. I just noticed an error in my chart, the 032 near the bottom is not all odd or all even, so the jump should not be colored green.

The next column is the numbers I was talking about. I need a term for those!! Someone give me suggestions, please!

Looking at the top of the chart, the number for tonight's draw is there. 05278-39164. Those are the 10 digits in order of draw. I number the positions as 12345-67890. To get that number I start with the last draw and write down the digits in order, but make sure that I do not duplicate. The last draw (04-14 mid) was 052 so my number starts with that. Next is 783. Since there is no duplication of the numbers that I have already used, I write those down, putting a dash after the #5 position which makes it easier to look at and determine the positions. Next is 533. I have already used 5 and 3, so I don't write them down again. Next is 091. I have already used the 0, but not the 9 and 1 so I write them down. Next is 200...you get the idea. I have found that most of the time the number drawn does not have all three digits coming from the same group of 5. I call them "Column A" and "Column B" because I used to post them in two columns. So I use an Excel spreadsheet to do that for me. It starts with 039, 031, 036, 034, 091, 096....etc. Using one from A and 2 from B, then it would use 2 from A and 1 from B.

When you do the A1B2, you will end up with 75 numbers. Another 75, of course, for A2B1. Too many to play!!

So, you have to decide how to cut down the total tickets. I do this in a number of different ways, but usually by looking at the positions of the drawn numbers. That is in the next column. You will see that the last draw, 052, had the 10-digit number of 78350-91264. That number was calculated as I just explained and was waiting for the draw to happen. After the draw, I entered the positions of the numbers that were drawn. So, the first digit was 0...that is position 5...the next digit was 5...that was position 4...the next digit was 2...that was position 8. So, that's how I got the "548" in that column. The next column is the distribution of numbers from Column A and Column B with doubles shown in "()".

Now, to try to figure out which number I'm going to use as a Key # (have that number in all my tickets), I use the column with the positions drawn in it. Here is what I wrote down on paper from the above to decide which number to use for the evening draw:

1=7 6=5

2=2 7=3

3=9 8=1

4=1 9=2

5=1 0=2

So, the first position numbers are 548. You will notice that in my chart 5, 4, and 8 have a "1". Then going back to the second draw back, I have 092, so 0, 9, and 2 are "2" in my chart. Then "(577)". I'm on draw 3...I already have a number for 5, so I put a "3" next to the 7 on my chart, etc. You should end up with the chart I just showed. Notice that the "3", which is position 3, has the largest number meaning that position has not been in the draw for 9 times. I will use the number that is in that position as my Key #. In this case, the real number 2 is in position 3 : 05278-39164. I now have the number I'm going to use as my Key #. I put the 10-number number (sure need a name for that!!) into my spreadsheet, it gives me all 150 combos and I go through and select only those that contain a "2". This gives me 40 tickets to play.

You asked how many to play. That's hard to answer. If you think the method works well enough for you and your state, then play them all as a box. If not, pick numbers that "look good" to you!!

Here is a count of the distributions of the columns for the draws I have in my database for Georgia:

(0-3) =

4

(1-2) =

6

(2-1) =

8

(3-0) =

3

0-3 =

7

1-2 =

25

2-1 =

25

3-0 =

9

This shows you that 1 from A and 2 from B and 2 from A and 1 from B are tied right now at 25 each. Again, the doubles are the ones in "()". So, the 0-3 count of 7 is for numbers that were not doubles, and the "(0-3)" count of 4 indicates that 4 times there were doubles with all three numbers from column B, or the second set of 5, whichever way you want to look at it.

Thanks, I really appreciate the help. I think I am understanding most of it, and will certainly go over it several more times until I do. However, if you play each of the 40 numbers as a box (50 cents each) and if you do get LUCKY and one hits, then you ONLY win $40 and it cost $20 to play (which doubling your money is always a good thing). However, my next question would be, how often has this method worked for you in an ongoing Lottery? Again, thanks for your help and any other help will be greatly appreciated from you or others!!!